Explanation Given: The rational inequality is given by 2 − 5 x + 2 x 2 ≥ 0 . Formula Used: When the inequality is of the form p ( x ) q ( x ) ≥ 0 then we have to solve p ( x ) = 0 a n d q ( x ) = 0 which is called as boundary value. We use the boundary value to separate the number line into disjoint interval. On each interval, p ( x ) q ( x ) is always positive or negative. We solve the inequality, either by making the table of testing the value for p ( x ) q ( x ) . Calculation: 2 − 5 x + 2 x 2 ≥ 0 ⇒ 2 x 2 − 4 x − x + 2 x 2 ≥ 0 ⇒ ( 2 x − 1 ) ( x − 2 ) x 2 ≥ 0 We have to set the numerator as well as denominator equal to 0 . Numerator:- ( x − 2 ) ( 2 x − 1 ) = 0 Denominator:- x 2 = 0 So, the boundary values are 0 , 1 2 a n d 2 which separate the number line into disjoint interval ( − ∞ , 0 , ( 0 , 1 2 ) , ( 1 2 , 2 ) a n d ( 2 , ∞ ) . When x = 1 2 o r 2 then p ( x ) q ( x ) = 0 , so 0 is neither positive nor negative. Since, due to equality sign we need to include where the inequality is 0 , so we can say that 1 2 a n d 2 is included in the interval. When x = 0 then p ( x ) q ( x ) = U n d e f i n e d , so 0 is excluded from the interval...

College Algebra with Modeling and Visualization with Mymathlab Access Code

5th Edition

ISBN: 9780321869418

Author: Gary K. Rockswold

Publisher: PEARSON

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Chapter 4.7, Problem 73E

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To solve the rational inequality.

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Chapter 4.7, Problem 72E

Chapter 4.7, Problem 74E

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College Algebra with Modeling and Visualization with Mymathlab Access Code

Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

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